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Conservation of energy

  1. Apr 1, 2012 #1
    1) [itex]\frac{1}{2}[/itex]mv[itex]^{2}_{2}[/itex]-mgy[itex]_{2}[/itex] = [itex]\frac{1}{2}[/itex]mv[itex]^{2}_{1}[/itex]-mgy[itex]_{1}[/itex]

    2) [itex]\frac{1}{2}[/itex]m(v[itex]^{2}_{2}[/itex]-2gy[itex]_{2}[/itex]) = [itex]\frac{1}{2}[/itex]m(v[itex]^{2}_{1}[/itex]-2gy[itex]_{1}[/itex])

    since g = [itex]\frac{GM}{R^{2}}[/itex] and [itex]\frac{1}{2}[/itex]m cancels.

    3) v[itex]^{2}_{2}[/itex]-2[itex]\frac{GM}{R^{2}}[/itex]y[itex]_{2}[/itex] = v[itex]^{2}_{1}[/itex]-2[itex]\frac{GM}{R^{2}}[/itex]y[itex]_{1}[/itex]

    4) v[itex]^{2}_{2}[/itex]-v[itex]^{2}_{1}[/itex] = 2[itex]\frac{GM}{R^{2}}[/itex]y[itex]_{2}[/itex]-2[itex]\frac{GM}{R^{2}}[/itex]y[itex]_{1}[/itex]

    5) v[itex]^{2}_{2}[/itex]-v[itex]^{2}_{1}[/itex] = 2[itex]\frac{GM}{R^{2}}[/itex](y[itex]_{2}[/itex]-y[itex]_{1}[/itex])

    6) [itex]\frac{Δ(v^{2})}{Δy}[/itex] = 2[itex]\frac{GM}{R^{2}}[/itex], if R is along y-axis

    then d(v[itex]^{2}[/itex]) = 2[itex]\frac{GM}{y^{2}}[/itex]dy

    Can someone give me a tip on where to go from here? Would an integral or derivative have any significance? Or are there any errors? Any feedback would be greatly appreciated. ( I know v = √2gy )
     
    Last edited: Apr 1, 2012
  2. jcsd
  3. Apr 2, 2012 #2

    Doc Al

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    Staff: Mentor

    Why the minus signs?

    Note that PE = mgy only holds close to the earth's surface. g is a constant and R here equals the radius of the earth. If you want to describe PE for larger ranges of distance, you cannot use PE = mgy. For a more general expression for potential energy see: Gravitational Potential Energy
     
  4. Apr 2, 2012 #3
    Well, doc, you might remember that I was banned for seven days for being stubborn that the potential energy is the work done to bring an object from infinity to the point? The same is said in the link you provided.
     
  5. Apr 2, 2012 #4

    Doc Al

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    Staff: Mentor

    There's nothing wrong with such a statement, when applied to situations where the force goes to zero at infinity.
     
  6. Apr 2, 2012 #5
    Isn't U(x) = -G(x)
     
  7. Apr 2, 2012 #6
    from the last part d(v[itex]^{2}[/itex]) = [itex]\frac{2GM}{y^{2}}[/itex]dy, if you take the integral, ∫d(v[itex]^{2}[/itex]) = ∫2[itex]\frac{GM}{y^{2}}[/itex]dy

    v[itex]^{2}[/itex] = -[itex]\frac{2GM}{y}[/itex], set v to c and rearrange y, you get

    y=-[itex]\frac{2GM}{c^{2}}[/itex]

    or

    R[itex]_{s}[/itex]=-[itex]\frac{2GM}{c^{2}}[/itex], would this be a valid derivation of the Schwarzschild radius?
     
  8. Apr 2, 2012 #7

    Doc Al

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    I don't understand your notation.

    U(x) = -∫F(x)dx
     
  9. Apr 2, 2012 #8
    Yeah G(x) =∫F(x)dx, thats what my teacher uses.
     
  10. Apr 2, 2012 #9
    I see why it is not a minus sign, because from the integral and and the minus sign, it will be positive.
     
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